hdu 3448(搜索+剪枝)Bag Problem

题意：

给n(n<=40)个物品和背包的容量w以及背包能装的物品个数k,每个物品有一个重量,问在满足背包的限制的情况下最多可以装多少物品。

思路

做过类似的题目,第一反应就是爆搜每个物品的两个状态放和不放。2^40肯定不行，来剪枝吧。
先把物品从小到大排序。一个有效的剪枝就是,最大的k个物品的重量和小于w那么这个重量和就是答案了,也是搜索中比较极限的情况,避免了去做搜索。
这样可以水过了,几乎没跑时间。5S有点吓人的意思?。
或许可以出可以卡这种剪枝的数据?

复杂度:

反正几乎没跑时间2333.

参考code:

/* #pragma warning (disable: 4786) #pragma comment (linker, "/STACK:0x800000") */#include <cassert>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <sstream>#include <iomanip>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <algorithm>#include <iterator>#include <utility>using namespace std;template< class T > T _abs(T n) { return (n < 0 ? -n : n); }template< class T > T _max(T a, T b) { return (!(a < b) ? a : b); }template< class T > T _min(T a, T b) { return (a < b ? a : b); }template< class T > T sq(T x) { return x * x; }template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }template< class T > bool inside(T a, T b, T c) { return a<=b && b<=c; }#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define repok(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))            #define MEM0(addr) memset((addr), 0, sizeof((addr)))#define MP(x, y) make_pair(x, y)#define REV(s, e) reverse(s, e)#define SET(p) memset(pair, -1, sizeof(p))#define CLR(p) memset(p, 0, sizeof(p))#define MEM(p, v) memset(p, v, sizeof(p))#define CPY(d, s) memcpy(d, s, sizeof(s))#define READ(f) freopen(f, "r", stdin)#define WRITE(f) freopen(f, "w", stdout)#define SZ(c) (int)c.size()#define PB(x) push_back(x)#define ff first#define ss second#define ll long long#define ld long double#define pii pair< int, int >#define psi pair< string, int >#define ls u << 1#define rs u << 1 | 1#define lson l, mid, u << 1#define rson mid, r, u << 1 | 1const int maxn = 1e6+5;const ll INF = 0x3f3f3f3f3f3fLL;ll n,w,k;ll a[45];ll ans;int vis[45];void dfs(int pos,int num,ll sum){    if(num > k || pos > n) return;    if(sum > ans) ans = sum;    rep(i,pos+1,n){        if(!vis[i] && a[i] + sum <= w){            vis[i] = 1;            dfs(pos + 1,num + 1,sum + a[i]);            vis[i] = 0;        }    }}int main(){    READ("in.txt");    while(scanf("%lld%lld",&k,&w)!=EOF){        scanf("%lld",&n);        rep(i,1,n) scanf("%lld",&a[i]);        sort(a+1,a+1+n);        ll sum = 0;        urep(i,n,n-k+1)            sum += a[i];        if(sum <= w){            printf("%lld\n", sum);            continue;        }        MEM0(vis);        ans = 0;        dfs(0,0,0);        printf("%lld\n", ans);    }    return 0;}